# Strange Printer LeetCode Solution

Here, We see Strange Printer LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

There is a strange printer with the following two special properties:

• The printer can only print a sequence of the same character each time.
• At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters.

Given a string `s`, return the minimum number of turns the printer needed to print it.

Example 1:
Input: s = “aaabbb”
Output: 2
Explanation: Print “aaa” first and then print “bbb”.

Example 2:
Input: s = “aba”
Output: 2
Explanation: Print “aaa” first and then print “b” from the second place of the string, which will cover the existing character ‘a’.

## Strange Printer LeetCode Solution C++

``````class Solution {
public:
int strangePrinter(string s) {
int n = s.size();
vector<vector<int>> dp(n, vector<int>(n, 0));

for (int i = n-1; i >= 0; --i) {
dp[i][i] = 1;
for (int j = i+1; j < n; ++j) {
dp[i][j] = dp[i][j-1] + 1;
for (int k = i; k < j; ++k) {
if (s[k] == s[j]) {
dp[i][j] = min(dp[i][j], dp[i][k] + (k+1<=j-1 ? dp[k+1][j-1] : 0));
}
}
}
}
return dp[0][n-1];
}
};```Code language: PHP (php)```

## Strange Printer LeetCode Solution Java

``````class Solution {
public int strangePrinter(String s) {
int n = s.length();
int[][] dp = new int[n][n];

for (int i = n-1; i >= 0; --i) {
dp[i][i] = 1;
for (int j = i+1; j < n; ++j) {
dp[i][j] = dp[i][j-1] + 1;
for (int k = i; k < j; ++k) {
if (s.charAt(k) == s.charAt(j)) {
dp[i][j] = Math.min(dp[i][j], dp[i][k] + (k+1<=j-1 ? dp[k+1][j-1] : 0));
}
}
}
}
return dp[0][n-1];
}
}```Code language: JavaScript (javascript)```

## Strange Printer Solution JavaScript

``````var strangePrinter = function(s) {
let n = s.length;
let dp = Array.from(Array(n), () => new Array(n).fill(0));
for (let i = n-1; i >= 0; --i) {
dp[i][i] = 1;
for (let j = i+1; j < n; ++j) {
dp[i][j] = dp[i][j-1] + 1;
for (let k = i; k < j; ++k) {
if (s[k] == s[j]) {
dp[i][j] = Math.min(dp[i][j], dp[i][k] + (k+1<=j-1 ? dp[k+1][j-1] : 0));
}
}
}
}
return dp[0][n-1];
};```Code language: JavaScript (javascript)```

## Strange Printer Solution Python

``````class Solution(object):
def strangePrinter(self, s):
n = len(s)
dp = [[0]*n for _ in range(n)]

for i in range(n-1, -1, -1):
dp[i][i] = 1
for j in range(i+1, n):
dp[i][j] = dp[i][j-1] + 1
for k in range(i, j):
if s[k] == s[j]:
dp[i][j] = min(dp[i][j], dp[i][k] + (dp[k+1][j-1] if k+1<=j-1 else 0))

return dp[0][n-1]``````
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